For which values of m does the function f(x) = x^m satisfy the equations a) 3x^2 d²y/dx² + 11x dy/dx + 3y = 0 and b) x² d²y/dx² + x dy/dx + 5y = 0?

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Determining Values of m for f(x) = x^m

To find the values of m for which the function f(x) = x^m is a solution to the given equations, we need to analyze each equation separately. We will perform this by substituting f(x) into each equation and simplifying to see under what conditions the equations hold true.

Equation a: 3x² d²y/dx² + 11x dy/dx + 3y = 0

First, we compute the first and second derivatives of f(x):

  • First derivative: dy/dx = mx^(m-1)
  • Second derivative: d²y/dx² = m(m-1)x^(m-2)

Substituting these into Equation a:

  3x²(m(m-1)x^(m-2)) + 11x(mx^(m-1)) + 3(x^m) = 0

Simplifying:

  3m(m-1)x^m + 11mx^m + 3x^m = 0

Factoring out x^m (assuming x is not equal to 0):

  x^m(3m(m-1) + 11m + 3) = 0

For this to hold, we need:

  3m(m-1) + 11m + 3 = 0

This simplifies to:

  3m² + 8m + 3 = 0

Using the quadratic formula:

  m = rac{-8 	extpm 	extsqrt{(8)² - 4(3)(3)}}{2(3)}

Calculating further, we find the two possible values for m.

Equation b: x² d²y/dx² + x dy/dx + 5y = 0

Next, we apply a similar approach to Equation b. Substituting the same derivatives:

  x²(m(m-1)x^(m-2)) + x(mx^(m-1)) + 5(x^m) = 0

Simplifying:

  m(m-1)x^m + mx^m + 5x^m = 0

Factoring out x^m gives:

  x^m(m(m-1) + m + 5) = 0

We need:

  m(m-1) + m + 5 = 0

This simplifies to:

  m² + 5 = 0

Again, applying the quadratic formula results in:

  m = 	extpm i 	extsqrt{5}

In summary, for:

  • Equation a: The values of m are determined from the quadratic equation 3m² + 8m + 3 = 0.
  • Equation b: The values of m are complex numbers given by m² + 5 = 0.

In conclusion, the conditions for m are contingent on the specific equation, where Equation a yields real solutions and Equation b yields complex solutions.

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